The base of a cubical tank is `25 m xx 40 m`. The volume of water in the tank is increasing at the rate of `500 m^(3)//min`. Find the rate at which the height of water is increasing.

Answer

To solve the problem step by step, we can follow this approach:

### Step 1: Understand the problem
We have a cubical tank with a base area of \(25 \, m \times 40 \, m\). The volume of water in the tank is increasing at a rate of \(500 \, m^3/min\). We need to find the rate at which the height of the water is increasing.

### Step 2: Define the variables
Let:
- \( V \) = volume of water in the tank (in \( m^3 \))
- \( h \) = height of water in the tank (in meters)
- \( A \) = area of the base of the tank = \( 25 \times 40 = 1000 \, m^2 \)
- \( \frac{dV}{dt} \) = rate of change of volume = \( 500 \, m^3/min \)
- \( \frac{dh}{dt} \) = rate of change of height (what we want to find)

### Step 3: Write the volume formula
The volume of water in the tank can be expressed as:
\[ V = A \cdot h = 1000 \cdot h \]

### Step 4: Differentiate with respect to time
Now, we differentiate both sides with respect to time \( t \):
\[ \frac{dV}{dt} = 1000 \cdot \frac{dh}{dt} \]

### Step 5: Substitute known values
We know that \( \frac{dV}{dt} = 500 \, m^3/min \). Substituting this into the equation gives:
\[ 500 = 1000 \cdot \frac{dh}{dt} \]

### Step 6: Solve for \( \frac{dh}{dt} \)
Now, we can solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{500}{1000} = \frac{1}{2} \, m/min \]

### Conclusion
The rate at which the height of the water is increasing is:
\[ \frac{dh}{dt} = \frac{1}{2} \, m/min \]

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